A reinforced concrete beam is deliberately designed to be under-reinforced because

steel yields before the concrete crushes, so failure is ductile and gives visible warning

the concrete crushes first, which gives the highest possible moment capacity for the steel provided

less concrete is required in the section, which is always the cheapest overall solution

no stirrups are then needed to resist shear in the beam near the supports

Prestressed concrete needs high-strength steel mainly because

creep, shrinkage and relaxation cost a tendon a nearly fixed strain, so a low-strain tendon loses most of its force

high-strength steel does not corrode at all, so the ducts need no grout and the tendons need no concrete cover at all

it bonds to concrete much better than any ordinary reinforcing bar can do, whatever the surface condition of the bar

the concrete is too weak to carry any compression unless such steel is present within the member itself, whatever its grade

creep, shrinkage and relaxation cost a tendon a nearly fixed strain, so a low-strain tendon loses most of its force

In a tall building, the main purpose of outriggers and a belt truss that connect the central core to the perimeter columns is to

make the core and perimeter columns resist overturning together, cutting core moment and drift

replace the core by carrying the floor gravity load from the slabs down to the foundation

let the building sway freely at the top so that the energy of the wind is absorbed

make the core and perimeter columns resist overturning together, cutting core moment and drift

add mass near the top of the building so that its natural period is shortened

The strong-column weak-beam principle in seismic design is followed so that

plastic hinges form in the beams, which dissipate energy without collapsing a storey

columns yield first and so give a visible warning before the beams are damaged at all

beams can be made much thinner, which lowers the seismic weight of the whole frame

the frame becomes much stiffer in every storey, so that its natural period is shortened

A thin hemispherical dome carries its own weight. The hoop (parallel-circle) stress in it is

compressive near the crown and tensile low down, which a base ring must resist

compressive everywhere in the shell, so that no base ring beam is needed at all

tensile everywhere in the shell, so that the dome must be built of reinforced concrete

zero everywhere, since a dome carries its load by the meridional force alone

compressive near the crown and tensile low down, which a base ring must resist

Which of the following best describes a well-conceived transitional shelter after a disaster?

A light, quickly erected shelter of local or reusable materials that is safe and can later be upgraded toward permanent housing

A permanent house of heavy masonry built as fast as possible by outside contractors so that no later reconstruction is needed

A short-life shelter made from materials that cannot be reused, so that they are never diverted to permanent housing later

A light, quickly erected shelter of local or reusable materials that is safe and can later be upgraded toward permanent housing

A shelter placed wherever the land is cheapest, since its location has little effect on the recovery of the people

Which statements about limit state design of reinforced concrete to IS 456 are correct?

The partial safety factors for materials are 1.5 for concrete and 1.15 for steel

The design stress of steel in flexure is taken as 0.87 fy

The ultimate compressive strain of concrete in flexure is taken as 0.0035

The partial safety factors for materials are 1.5 for concrete and 1.15 for steel

The design stress of steel in flexure is taken as 0.87 fy

The ultimate compressive strain of concrete in flexure is taken as 0.0035

Serviceability limit states are checked at the factored load 1.5(DL + IL)

Which statements about lateral load-resisting systems for tall buildings are correct?

A shear wall core is very stiff in bending and carries most of the lateral shear

A moment frame gets its lateral stiffness from rigid beam-column joints

A framed tube makes the perimeter act like a hollow cantilever, with shear lag reducing its efficiency

A shear wall core is very stiff in bending and carries most of the lateral shear

A moment frame gets its lateral stiffness from rigid beam-column joints

A framed tube makes the perimeter act like a hollow cantilever, with shear lag reducing its efficiency

Lateral drift is controlled mainly by adding weight to the upper floors

Which of the following increase the seismic vulnerability of a building?

An open ground storey for parking under upper floors with masonry infill

Short columns created by infill walls that stop below the beam

Re-entrant corners that make the plan L-shaped and torsion-prone

An open ground storey for parking under upper floors with masonry infill

Short columns created by infill walls that stop below the beam

A regular, symmetric plan with uniform stiffness over the height

Re-entrant corners that make the plan L-shaped and torsion-prone

Which statements about arches, cables and space frames are correct?

A parabolic arch under a UDL on its horizontal span carries it in pure compression

A cable under a UDL on its horizontal span hangs as a parabola and carries only tension

The members of a space frame carry load essentially by axial tension or compression

A parabolic arch under a UDL on its horizontal span carries it in pure compression

A cable under a UDL on its horizontal span hangs as a parabola and carries only tension

Increasing the sag of a cable increases its horizontal tension

The members of a space frame carry load essentially by axial tension or compression

Structural Systems: Statics, Limit State Design, RCC and Steel, Prestressing, High-Rise, Shells, Long-Span and Disaster-Resistant Structures

This chapter answers the last part of AR Section B1.3: “Building structural systems; Elastic and Limit State design; Structural systems; Design with steel and RCC; Principles of Prestressing; High Rise and Shell structures; Gravity and lateral load resisting systems; Space frames; Long span structures; Suspension structures; Catenary structures; Wind load & seismic considerations in high-rise structures; Disaster resistant structures; Temporary structures for rehabilitation”. It teaches structural systems and the load path, beam statics and stress, working-stress and limit state design, the principles of RCC and steel design, prestressing, gravity and lateral systems of high-rise buildings with wind and seismic effects, shells, space frames, long-span, suspension and catenary structures, disaster-resistant design and temporary shelters. The numericals are beam moments and stresses, load takedown, column capacity, prestress stresses, base shear and its distribution, and arch and cable forces.

1. Structural systems, load path and basic statics

A structural system is the arrangement of elements that carries loads to the ground. The main families are the load-bearing wall, the frame (post-and-beam or rigid), the truss, the arch, the shell and folded plate, cable and membrane structures, the space frame, and the tube. Loads are dead (self-weight, finishes, fixed equipment), imposed or live (occupancy), wind, earthquake, and in some places snow. The load path runs slab to beam to girder to column to footing to soil, and every link must be strong and connected. Gravity loads act down; wind and earthquake are lateral loads that need a separate system as buildings grow taller.

Standard results for beams (span L, UDL w, point load P)
Case	Reaction / shear	Maximum moment	Maximum deflection
Simply supported, UDL	wL/2 at each support	wL²/8 at midspan	5wL⁴/(384EI)
Simply supported, central P	P/2 at each support	PL/4 at midspan	PL³/(48EI)
Cantilever, UDL	wL at the fixed end	wL²/2 at the fixed end	wL⁴/(8EI) at the tip
Cantilever, tip P	P	PL at the fixed end	PL³/(3EI)

Stress. Axial stress is σ = P/A; bending stress is σ = M/Z, and for a rectangle Z = bd²/6; Hooke's law σ = Eε and elongation δ = PL/AE. Worked example 1. A simply supported beam of 6 m span carries w = 20 kN/m: R = 60 kN, M = 20 × 36/8 = 90 kN·m. For a 300 × 600 mm section Z = 300 × 600²/6 = 18 × 10⁶ mm³, so σ = 90 × 10⁶/18 × 10⁶ = 5 MPa. Worked example 2, load takedown. A column carries the load of its tributary area, the plan area closer to it than to any other column, from every floor above. On a 6 m × 5 m grid the interior column has 30 m²; with five floors and a roof at 12 kN/m² each, it carries 6 × 30 × 12 = 2160 kN at the base, and on 400 × 400 mm that is 2160 × 10³/160 000 = 13.5 MPa.

⚠️ Cantilever versus simply supported

For the same span and UDL a cantilever's maximum moment, wL²/2, is four times a simply supported beam's wL²/8, and its tip deflection wL⁴/8EI is about ten times as large as 5wL⁴/384EI. Do not mix the PL/4 of a central load with the wL²/8 of a UDL.

2. Elastic (working stress) and limit state design

In the working stress method (WSM) materials are assumed elastic, stresses under service loads are kept below permissible stresses (a fraction of strength), and a single factor of safety sits on the materials. It is simple, but it does not show the real margin against collapse and treats uncertain live load like certain dead load. Limit state design (LSD) designs each member for stated limit states: ultimate (strength in flexure, shear, compression and torsion; stability, overturning) and serviceability (deflection, cracking, durability, vibration). Characteristic loads are multiplied by partial factors γf and characteristic strengths divided by partial factors γm.

WSM and LSD compared
Aspect	Working stress	Limit state
Basis	Elastic behaviour under service load	Behaviour at failure and in service, both checked
Safety	One factor on material strength	Partial factors on loads and on materials
Result	Margin against collapse unknown	More uniform reliability; ductility recognised

For RCC to IS 456, γm = 1.5 for concrete and 1.15 for steel, so the design strength of concrete in the stress block is 0.67 fck/1.5 = 0.446 fck and the design stress of steel is fy/1.15 = 0.87 fy. The load factors are 1.5 (DL + IL), 1.5 (DL ± WL or EL), 1.2 (DL + IL ± WL or EL), and 0.9 DL + 1.5 WL where overturning or uplift is checked. The ultimate concrete strain in flexure is 0.0035. Serviceability is checked at service (characteristic) loads. Steel structures to IS 800:2007 use LSD too, with γm0 = 1.10 on yield and γm1 = 1.25 on ultimate strength. Example. DL = 20 kN/m and LL = 12 kN/m on a 6 m simply supported beam give wu = 1.5 × 32 = 48 kN/m and Mu = 48 × 36/8 = 216 kN·m.

⚠️ Factored loads are for strength, not for deflection

Do not apply 1.5 when checking deflection or cracking: serviceability uses service loads. And do not divide concrete's 0.67 fck by 1.5 twice or forget it: the stress-block value is 0.446 fck.

3. Design with RCC and steel: principles

RCC works because concrete is strong in compression and weak in tension (roughly a tenth of its compressive strength), steel carries the tension, the two bond together, expand almost equally with heat, and the cover protects the steel. Beams are designed for flexure, shear (stirrups) and deflection. A section is under-reinforced when steel yields before the concrete crushes (ductile, with warning) and over-reinforced when concrete crushes first (brittle), and designers always aim for the first; the limiting neutral-axis depth xu,max/d is 0.53, 0.48 and 0.46 for Fe 250, Fe 415 and Fe 500, and for Fe 415 the limiting moment is Mu,lim = 0.138 fck b d². Moment capacity of an under-reinforced rectangular section is Mu = 0.87 fy Ast d [1 − Ast fy/(b d fck)]. For b = 250, d = 450 mm, Ast = 804 mm² (4 bars of 16 mm), fy = 415 and fck = 20 MPa, Mu = 111.3 kN·m. Minimum tension steel in a beam is 0.85bd/fy, and the maximum is 4 per cent of bD.

Slabs span one way if the long-to-short span ratio exceeds 2 and two ways if it is 2 or less; flat slabs on columns need drops or capitals against punching shear. Columns carry axial load and moment; a short column's axial capacity is Pu = 0.4 fck Ac + 0.67 fy Asc. For a 300 × 300 mm column of M20 and Fe 415 with four 20 mm bars (Asc = 4 × 314.2 = 1256.6 mm², Ac = 88 743 mm²), Pu = 709.9 + 349.4 = 1059 kN. Longitudinal steel is 0.8 to 6 per cent of the gross area, with at least four bars (six in a circular column) of at least 12 mm; lateral ties confine the core. Nominal cover (IS 456) is 20 mm for mild exposure, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme, with at least 40 mm to column bars and 50 mm in footings.

Steel design (IS 800 limit state) treats tension members by yielding on the gross section, Tdg = Ag fy/γm0, and compression members by buckling, governed by slenderness KL/r. Effective-length factors are K = 1 (both ends pinned), 0.5 (both fixed), 0.7 (one fixed, one pinned) and 2 (one fixed, one free), and the Euler load is Pcr = π²EI/(KL)²: for E = 200 GPa, I = 8 × 10⁶ mm⁴ and a pinned 4 m length, Pcr = 987 kN. Sections are plastic, compact, semi-compact or slender by their resistance to local buckling. A laterally restrained beam can reach its full moment capacity; an unrestrained one fails earlier by lateral-torsional buckling. The plastic moment is Mp = fy Zp, and the shape factor Zp/Ze is 1.5 for a rectangle and about 1.1 to 1.2 for rolled I-sections.

4. Principles of prestressing

Concrete cracks in tension long before steel yields. Prestressing puts the concrete into compression before the loads act, so that load-induced tension first cancels the precompression and the member stays uncracked, uses its whole section, is stiffer, and can span farther or be shallower. Stress at a fibre is σ = P/A ± Pe·y/I ± M·y/I. Example. A 300 × 600 mm beam with P = 1200 kN at e = 100 mm below the centroid has P/A = 1200 × 10³/180 000 = 6.67 MPa and Pe/Z = 1200 × 10³ × 100/(18 × 10⁶) = 6.67 MPa, so the bottom fibre is at 13.33 MPa and the top is at zero: e = h/6 = 100 mm is the limit of the kern, within which a prestress leaves the whole section in compression.

Pre-tensioning and post-tensioning
Aspect	Pre-tensioning	Post-tensioning
Sequence	Tendons are stressed between anchors, then concrete is cast and the tendons are released	Concrete is cast with ducts, hardens, then tendons are stressed against the member
Force transfer	By bond along the tendon	By end anchorages, and bond if grouted
Place and shape	Factory, straight or lightly deflected tendons; hollow-core planks, sleepers, piles	Site or factory, any curved profile; bridges, slabs, transfer girders
Losses	Elastic shortening, creep, shrinkage, relaxation; no friction loss	Also friction between tendon and duct and anchorage slip

Losses reduce the initial prestress: elastic shortening, anchorage slip, friction (post-tensioning only), creep and shrinkage of concrete, and relaxation of steel; a total of about 15 to 20 per cent is a rough allowance, and an initial 1000 kN at 15 per cent loss leaves 850 kN. This is why prestressing needs high-strength steel and high-strength concrete: creep, shrinkage and relaxation shorten the tendon by almost fixed strains, so an ordinary steel stressed at low strain would lose nearly all of its force, while a highly strained tendon keeps most of it. Load balancing gives a clear picture: a parabolic tendon with sag e in span L exerts an upward load w = 8Pe/L², so P = 1000 kN, e = 0.15 m and L = 10 m balance 12 kN/m of downward load.

⚠️ Why not ordinary steel

Using mild steel for a prestressing tendon fails because its working strain is far below the strain lost to creep and shrinkage. A second slip: friction loss belongs to post-tensioning only, because a pre-tensioned tendon is not in a duct.

5. High-rise structures: gravity and lateral load-resisting systems

In a tall building gravity load grows in proportion to height, but lateral load effects grow much faster. Under a uniform wind load w per unit height a building acts as a vertical cantilever: base shear is wH, base moment wH²/2 and top deflection wH⁴/8EI. Doubling the height at the same w and EI therefore multiplies the base moment by 4 and the deflection by 16. For w = 20 kN/m on a 60 m building, base shear is 1200 kN and base moment 36 000 kN·m. So stiffness and drift (commonly kept near H/500 under wind) govern the design of tall buildings, not strength alone.

Lateral load-resisting systems
System	How it resists	Remarks
Rigid (moment-resisting) frame	Rigid beam-column joints bend the members	Open plan, flexible, suits low and mid-rise; drift mainly by storey shear
Shear wall (often the lift or stair core)	Deep RC wall acts as a vertical cantilever	Very stiff and economical; limits planning; bends at the base
Dual (frame plus shear wall)	Floor slabs force wall and frame to deflect together	Wall controls the lower storeys, frame the upper; stiffer than either alone
Braced frame (X, K, V, eccentric)	Diagonals turn lateral load into axial forces	Efficient in steel; eccentric bracing adds ductility
Outrigger and belt truss	Stiff arms tie the core to perimeter columns so they resist overturning together	Cuts core moment and drift; the outrigger floors cost usable space
Tube (framed, trussed, bundled, tube-in-tube) and diagrid	The perimeter acts as a hollow cantilever box; diagrid diagonals carry gravity and lateral load	Very tall; shear lag leaves corner columns more stressed than middle ones; tube systems are associated with Fazlur Khan

🎯 The floor is a diaphragm

Whatever the system, floor slabs act as rigid horizontal diaphragms that collect wind and inertia loads and pass them to the walls, braces or frames. A building with big floor openings or a weak diaphragm cannot share lateral load among its vertical elements, whatever their strength.

6. Wind load and seismic considerations in high-rise structures

Wind. The design wind speed is Vz = Vb·k₁·k₂·k₃·k₄, the basic speed modified for risk, terrain and height, topography and importance, and the design pressure is pz = 0.6 Vz² (N/m², Vz in m/s): for Vz = 40 m/s, pz = 960 N/m². Speed and pressure rise with height and fall with rough terrain. Slender towers feel across-wind vortex-shedding response as well as along-wind push, and flexible ones need dynamic analysis with a gust factor; comfort (acceleration at the top), not strength, often governs. The remedies are aerodynamic shaping (tapering, setbacks, rounded or chamfered corners, through-building openings), more stiffness, damping, and tuned mass dampers such as the large pendulum mass in Taipei 101, with wind-tunnel tests for important towers.

Earthquake. The inertia force is mass times ground acceleration, so seismic design cares about mass, stiffness, period and ductility. In IS 1893 (Part 1), 2016 edition, the design horizontal coefficient is Ah = (Z/2)(I/R)(Sa/g) and the design base shear is VB = Ah W, where Z = 0.10, 0.16, 0.24 or 0.36 for zones II to V of that edition, I is the importance factor, R the response-reduction factor (higher for ductile systems) and W the seismic weight, being dead load plus 25 per cent of imposed load up to 3 kN/m² or 50 per cent above it. For Z = 0.36, I = 1, R = 5 and Sa/g = 2.5, Ah = 0.18 × 0.2 × 2.5 = 0.09, so a weight of 20 000 kN gives VB = 1800 kN. The base shear is distributed over the height by Qi = VB·Wi·hi²/Σ Wj·hj²: for equal floor weights at 3, 6 and 9 m and VB = 1260 kN, the top floor gets 1260 × 81/126 = 810 kN. The approximate period of a bare RC frame is Ta = 0.075h^0.75 (h in m), and of a steel frame 0.085h^0.75.

Configuration: regular plan and elevation, symmetry, a continuous load path and uniform stiffness. Avoid a soft storey (open ground floor under stiff infilled floors), short columns formed by partial-height infill, re-entrant corners and torsion, and floating columns; keep adjacent buildings apart by a seismic gap against pounding.
Ductility: strong column, weak beam, so that plastic hinges form in beams that can dissipate energy without collapsing a storey; closely spaced ties with 135° hooks at joints and member ends (IS 13920). Base isolation (lead-rubber or friction-pendulum bearings) lengthens the period and cuts the force; dampers dissipate energy.

⚠️ Not every load rises with height the same way

Wind pressure rises with height and acts on exposed area, so it governs slender, flexible towers; earthquake force grows with mass, so it governs heavy, stiff, low and mid-rise buildings. Making upper floors lighter helps in an earthquake but does little against wind.

7. Shells, space frames, long-span, suspension and catenary structures

A shell is a thin curved surface that carries load mainly by in-plane membrane forces (compression, tension and shear) with little bending, so a very thin concrete shell can span far. Its edges need support or a ring or edge beam to take the thrust. By curvature: singly curved shells (cylindrical barrel vaults, cones) are curved in one direction; doubly curved synclastic shells (domes, elliptic paraboloids) curve the same way in both directions; anticlastic shells (hyperbolic paraboloid or hypar, hyperboloid) curve oppositely, and since hypars and hyperboloids are made of straight lines their formwork is straight timber. In a dome under self-weight the meridional force is compressive everywhere, while the hoop force is compressive near the crown and turns to tension below about 52° from the crown, which the base ring beam must resist. A folded plate spans like a deep beam across its folds. Examples include Candela's hypar shells, Nervi's ribbed concrete vaults, and the Sydney Opera House shells.

A space frame is a three-dimensional truss, usually a double-layer grid of top and bottom chords joined by diagonals at modular nodes (ball-type joints); its members carry axial tension or compression only, it spreads load in two or more directions, and a shallow depth spans large column-free areas. A truss is determinate and stable when m + r = 2j in a plane and m + r = 3j in space (m members, r reaction components, j joints): a plane truss with 8 joints and 3 reactions has 13 members. Other long-span forms are trusses (Warren, Pratt, Vierendeel with rigid joints and no diagonals), portal frames, lamella and geodesic domes, and arches: a three-hinged parabolic arch under a UDL over its horizontal span carries it in pure compression with horizontal thrust H = wL²/8f (f the rise), so w = 20 kN/m, L = 24 m and f = 6 m give 240 kN, a thrust that abutments or a tie must resist.

Cables carry load in pure tension and take the shape of the load: a free-hanging cable under its own weight forms a catenary, while under a uniform load over its horizontal span it is a parabola, and the inverted catenary is the ideal arch for self-weight (the Gateway Arch is a flattened one). For span L, sag d and load w per horizontal metre, H = wL²/8d and the maximum tension at a support is Tmax = √(H² + (wL/2)²); a larger sag gives a smaller H. For w = 5 kN/m, L = 40 m and d = 4 m, H = 250 kN, V = 100 kN and Tmax = 269.3 kN. A suspension structure hangs a deck from main cables over towers by hangers, with anchorages to resist the cable pull and a stiffening girder or truss against wind flutter (the lesson of the 1940 Tacoma Narrows bridge); a cable-stayed structure runs straight stays from towers to the deck and is stiffer. Tensile membrane and cable-net roofs, as in Frei Otto's Munich Olympic roof, must be prestressed so they never go slack, and catenary roofs include Saarinen's Dulles terminal.

8. Disaster-resistant structures and temporary structures for rehabilitation

Disaster resistance starts with the hazard and the site, using hazard maps (India's seismic zone map, wind and cyclone, and flood-hazard maps) to avoid floodplains, unstable slopes and soft soils. The common principles are a simple, regular, symmetric form, a continuous load path, ductility, redundancy and good connections, and quality of materials and work.

Hazards and responses
Hazard	Key measures
Earthquake, masonry	Box action: plinth, lintel, roof and gable bands; vertical bars at corners and jambs; limited and well-placed openings; light roof; good mortar (IS 4326)
Earthquake, frames	Strong column, weak beam; ductile detailing; no soft storey, short column or floating column; seismic gaps; base isolation and dampers where justified
Cyclone and high wind	Hip rather than gable roofs, limited overhangs, roofs strapped down to walls, shuttered or small openings, protection from flying debris; raised shelters
Flood and tsunami	Raised plinth or stilts, open or breakaway ground floor, water-resistant materials, elevated services; vertical evacuation platforms
Landslide	Avoid unstable slopes, drain surface water, retain slopes, light construction on cuts

Temporary structures for rehabilitation serve people displaced by disaster. The sequence runs from emergency shelter (tents, tarpaulins; days to weeks), to temporary or transitional shelter (months to a few years, such as bamboo or sheet-metal frames, container and prefabricated units), to permanent housing. A good temporary structure is quick to deploy, light and transportable (often flat-pack), cheap, built from local or reusable materials with local labour, modular and demountable, safe against wind, flood and fire, comfortable for the climate, respectful of privacy and culture, and, in the best case, upgradable or reusable toward permanent housing rather than discarded. Structurally it is stabilised by triangulation or bracing and by ground anchors against wind uplift. Examples: tensile-fabric tents and A-frames, geodesic domes, bamboo frames, Shigeru Ban's paper-tube shelters and the Christchurch cardboard cathedral, shipping containers, and the Bailey bridge, a standard modular steel truss bridge for rapid reconnection.

⚠️ Temporary is not careless

A temporary shelter still has to survive the next wind, flood or winter and keep fire from spreading, and a camp laid out without spacing, drainage and access turns a disaster into a second one. Quick does not mean unengineered.

Key takeaways

Beam results to know: simply supported UDL M = wL²/8, central load M = PL/4, cantilever UDL M = wL²/2; σ = P/A, σ = M/Z with Z = bd²/6; a column carries its tributary area times the number of floors.
Limit state design uses partial factors (1.5 and 1.15 on materials; 1.5 on DL + IL), design strengths 0.446 fck and 0.87 fy, and aims for under-reinforced, ductile failure; serviceability is checked at service loads.
Prestressing precompresses concrete: σ = P/A ± Pe/Z, high-strength steel is essential because of fixed strain losses, friction loss is only in post-tensioning, and a parabolic tendon balances w = 8Pe/L².
In tall buildings, drift grows as H⁴ and base moment as H², so stiffness governs: shear walls, dual systems, outriggers, tubes and diagrids, with floors acting as diaphragms.
Wind pressure is pz = 0.6 Vz²; seismic base shear is VB = Ah W with Ah = (Z/2)(I/R)(Sa/g), distributed in proportion to Wi hi²; regular form, strong column-weak beam and ductile detailing are the seismic essentials.
Shells carry load by membrane action, space frames by axial members (m + r = 3j), arches in compression with H = wL²/8f, and cables in tension with H = wL²/8d; temporary shelters should be quick, safe, light, local and upgradable.

Practice questions (20)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For the same span and the same uniformly distributed load, the ratio of the maximum bending moment of a cantilever to that of a simply supported beam is
1. 2
2. 3
3. 4
4. 8
Show answer
Answer: C — 4
The cantilever's moment at the fixed end is wL²/2 and the simply supported beam's at midspan is wL²/8, so the ratio is (1/2)/(1/8) = 4. Using wL²/8 for both beams would give a ratio of 1, which is the common slip, and 2, 3 and 8 do not follow from the standard formulas.
In limit state design of reinforced concrete to IS 456, the design compressive strength used in the stress block, with γm = 1.5, is
1. 0.670 fck
2. 0.446 fck
3. 0.870 fck
4. 0.360 fck
Show answer
Answer: B — 0.446 fck
The 0.67 factor allows for the difference between cube and in-situ strength, and the material factor divides it by 1.5: 0.67/1.5 = 0.446 fck. The figure 0.67 fck, the tempting one, is before dividing by γm; 0.87 is the steel factor fy/1.15, and 0.36 appears in the stress block's compressive force 0.36 fck b xu, not as a design strength.
A reinforced concrete beam is deliberately designed to be under-reinforced because
1. steel yields before the concrete crushes, so failure is ductile and gives visible warning
2. the concrete crushes first, which gives the highest possible moment capacity for the steel provided
3. less concrete is required in the section, which is always the cheapest overall solution
4. no stirrups are then needed to resist shear in the beam near the supports
Show answer
Answer: A — steel yields before the concrete crushes, so failure is ductile and gives visible warning
Yielding steel stretches, the beam cracks and deflects visibly, and failure is gradual, whereas crushing of concrete is sudden and brittle. High capacity from an over-reinforced section is the tempting claim, but that capacity is unreliable because failure gives no warning. Shear reinforcement is needed in either case.
Prestressed concrete needs high-strength steel mainly because
1. high-strength steel does not corrode at all, so the ducts need no grout and the tendons need no concrete cover at all
2. it bonds to concrete much better than any ordinary reinforcing bar can do, whatever the surface condition of the bar
3. the concrete is too weak to carry any compression unless such steel is present within the member itself, whatever its grade
4. creep, shrinkage and relaxation cost a tendon a nearly fixed strain, so a low-strain tendon loses most of its force
Show answer
Answer: D — creep, shrinkage and relaxation cost a tendon a nearly fixed strain, so a low-strain tendon loses most of its force
The strain lost to creep, shrinkage and relaxation is of fixed size, so a tendon stressed to a low strain, like mild steel, would lose almost all its prestress, and a highly strained one retains most. High-strength steel is not immune to corrosion, which rules out the corrosion-immunity claim, and prestressed concrete is itself high-strength, so the weak-concrete claim is also wrong.
In a tall building, the main purpose of outriggers and a belt truss that connect the central core to the perimeter columns is to
1. replace the core by carrying the floor gravity load from the slabs down to the foundation
2. let the building sway freely at the top so that the energy of the wind is absorbed
3. make the core and perimeter columns resist overturning together, cutting core moment and drift
4. add mass near the top of the building so that its natural period is shortened
Show answer
Answer: C — make the core and perimeter columns resist overturning together, cutting core moment and drift
The stiff outrigger turns the column pull and push into a restoring couple, so the core's overturning moment and the drift fall. Replacing the core's gravity role is wrong because the outrigger is a lateral element. Allowing free sway is the opposite of the aim, and added mass at the top lengthens the period, not shortens it.
A tall building is idealised as a vertical cantilever under a uniform lateral load per unit height. If the height is doubled with the same load per unit height and the same flexural stiffness EI, the top deflection becomes
1. 4 times
2. 8 times
3. 16 times
4. 32 times
Show answer
Answer: C — 16 times
The tip deflection is wH⁴/8EI, so doubling H multiplies it by 2⁴ = 16. Four times is the tempting figure because the base moment wH²/2 rises by 2² = 4; deflection depends on a higher power of height than moment does.
The strong-column weak-beam principle in seismic design is followed so that
1. plastic hinges form in the beams, which dissipate energy without collapsing a storey
2. columns yield first and so give a visible warning before the beams are damaged at all
3. beams can be made much thinner, which lowers the seismic weight of the whole frame
4. the frame becomes much stiffer in every storey, so that its natural period is shortened
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Answer: A — plastic hinges form in the beams, which dissipate energy without collapsing a storey
If the columns yielded first, a hinge at the top and bottom of a storey's columns would form a storey mechanism and the building would collapse. Hinges in beams spread the damage and dissipate energy while the columns still carry gravity. The warning-from-columns idea is the tempting error, because column failure is catastrophic, not a warning.
A thin hemispherical dome carries its own weight. The hoop (parallel-circle) stress in it is
1. compressive everywhere in the shell, so that no base ring beam is needed at all
2. tensile everywhere in the shell, so that the dome must be built of reinforced concrete
3. zero everywhere, since a dome carries its load by the meridional force alone
4. compressive near the crown and tensile low down, which a base ring must resist
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Answer: D — compressive near the crown and tensile low down, which a base ring must resist
Near the crown the hoop force is compressive, and below about 52° from the crown it turns to tension, because the lower parts tend to spread outward; the ring beam at the base takes that tension. Compression everywhere is the tempting answer for a masonry dome, and it is why unreinforced masonry domes crack along meridians.
A three-hinged parabolic arch of span 24 m and rise 6 m carries a uniformly distributed load of 20 kN/m over its horizontal span. The horizontal thrust at the supports is
1. 120 kN
2. 240 kN
3. 480 kN
4. 960 kN
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Answer: B — 240 kN
H = wL²/8f = 20 × 24²/(8 × 6) = 20 × 576/48 = 240 kN. The figure 480 kN comes from using wL²/4f, which is a common slip when the 8 in the formula is forgotten, and 960 kN from wL²/2f.
Which of the following best describes a well-conceived transitional shelter after a disaster?
1. A permanent house of heavy masonry built as fast as possible by outside contractors so that no later reconstruction is needed
2. A short-life shelter made from materials that cannot be reused, so that they are never diverted to permanent housing later
3. A light, quickly erected shelter of local or reusable materials that is safe and can later be upgraded toward permanent housing
4. A shelter placed wherever the land is cheapest, since its location has little effect on the recovery of the people
Show answer
Answer: C — A light, quickly erected shelter of local or reusable materials that is safe and can later be upgraded toward permanent housing
Speed, safety, local or reusable materials and a route toward permanent housing are the defining features of a good transitional shelter. Heavy permanent masonry built in haste is the tempting wrong choice, since it cannot be quick, safe and affordable together, and a poor location raises risk and slows recovery.
Which statements about limit state design of reinforced concrete to IS 456 are correct?
1. The partial safety factors for materials are 1.5 for concrete and 1.15 for steel
2. The design stress of steel in flexure is taken as 0.87 fy
3. The ultimate compressive strain of concrete in flexure is taken as 0.0035
4. Serviceability limit states are checked at the factored load 1.5(DL + IL)
Show answer
Answer: A — The partial safety factors for materials are 1.5 for concrete and 1.15 for steel; B — The design stress of steel in flexure is taken as 0.87 fy; C — The ultimate compressive strain of concrete in flexure is taken as 0.0035
The three correct statements are the standard IS 456 values. Serviceability, meaning deflection and cracking, is checked at service loads rather than at the factored load, which is the tempting error because the factored load is used for the strength limit state.
Which statements about lateral load-resisting systems for tall buildings are correct?
1. A shear wall core is very stiff in bending and carries most of the lateral shear
2. A moment frame gets its lateral stiffness from rigid beam-column joints
3. A framed tube makes the perimeter act like a hollow cantilever, with shear lag reducing its efficiency
4. Lateral drift is controlled mainly by adding weight to the upper floors
Show answer
Answer: A — A shear wall core is very stiff in bending and carries most of the lateral shear; B — A moment frame gets its lateral stiffness from rigid beam-column joints; C — A framed tube makes the perimeter act like a hollow cantilever, with shear lag reducing its efficiency
Walls, moment frames and tubes are standard lateral systems, and tube efficiency is reduced because corner columns attract more load than middle ones (shear lag). Adding weight to the upper floors is the tempting wrong item: it raises inertia forces and does not add stiffness.
Which of the following increase the seismic vulnerability of a building?
1. An open ground storey for parking under upper floors with masonry infill
2. Short columns created by infill walls that stop below the beam
3. A regular, symmetric plan with uniform stiffness over the height
4. Re-entrant corners that make the plan L-shaped and torsion-prone
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Answer: A — An open ground storey for parking under upper floors with masonry infill; B — Short columns created by infill walls that stop below the beam; D — Re-entrant corners that make the plan L-shaped and torsion-prone
A soft storey, short columns and re-entrant corners all concentrate damage or twisting. A regular symmetric plan with uniform stiffness is the recommended configuration, and so reduces vulnerability.
Which statements about arches, cables and space frames are correct?
1. A parabolic arch under a UDL on its horizontal span carries it in pure compression
2. A cable under a UDL on its horizontal span hangs as a parabola and carries only tension
3. Increasing the sag of a cable increases its horizontal tension
4. The members of a space frame carry load essentially by axial tension or compression
Show answer
Answer: A — A parabolic arch under a UDL on its horizontal span carries it in pure compression; B — A cable under a UDL on its horizontal span hangs as a parabola and carries only tension; D — The members of a space frame carry load essentially by axial tension or compression
The arch and the cable are mirror images, one in compression and one in tension, and a space frame is a pin-jointed assembly of axial members. The sag statement is reversed because H = wL²/8d falls as the sag d rises, which is the tempting error since a deeper cable looks more heavily stressed.
A six-storey building (including the roof) has columns on a 6 m × 5 m grid. Each floor and the roof carry a total load of 12 kN/m². Ignoring column self-weight and live-load reduction, the axial stress in the ground-storey interior column of section 400 mm × 400 mm, in MPa, is ____ (one decimal place).
Numerical answer — type the value.
Show answer
Answer: 13.5
Tributary area = 6 × 5 = 30 m². Load = 6 floors × 30 × 12 = 2160 kN. Area = 400 × 400 = 160 000 mm². Stress = 2160 × 10³/160 000 = 13.5 MPa. Check: 13.5 × 160 000 = 2.16 × 10⁶ N.
A simply supported beam of span 6 m carries a dead load of 20 kN/m (including self-weight) and a live load of 12 kN/m. Using a load factor of 1.5 on the combination (DL + LL), the design maximum bending moment in kN·m is ____.
Numerical answer — type the value.
Show answer
Answer: 216
Factored load wu = 1.5 × (20 + 12) = 48 kN/m. Mu = wu L²/8 = 48 × 36/8 = 216 kN·m. Check: unfactored M = 32 × 36/8 = 144 kN·m and 1.5 × 144 = 216.
A 300 mm × 300 mm short RCC column of M20 concrete and Fe 415 steel has four 20 mm bars. Using Pu = 0.4 fck Ac + 0.67 fy Asc, where Ac is the gross area less Asc, the axial capacity in kN, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 1059
Asc = 4 × π × 20²/4 = 1256.6 mm². Ac = 90 000 − 1256.6 = 88 743.4 mm². Concrete term = 0.4 × 20 × 88 743.4 = 709 947 N. Steel term = 0.67 × 415 × 1256.6 = 349 409 N. Sum = 1 059 356 N = 1059 kN to the nearest integer.
A prestressed rectangular beam 300 mm wide and 600 mm deep carries an effective prestress of 1200 kN at an eccentricity of 100 mm below the centroid. Ignoring self-weight and external load, the stress at the bottom fibre, in MPa (compression positive), to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 13.33
A = 300 × 600 = 180 000 mm² and Z = 300 × 600²/6 = 18 × 10⁶ mm³. P/A = 1200 × 10³/180 000 = 6.667 MPa; Pe/Z = 1200 × 10³ × 100/(18 × 10⁶) = 6.667 MPa. The prestress is below the centroid, so the bottom fibre is compressed by both: 13.33 MPa. Check: top fibre = 6.667 − 6.667 = 0.
A three-storey building has equal floor weights at heights 3 m, 6 m and 9 m above the base, and a design base shear of 1260 kN. Using Qi = VB Wi hi²/Σ Wj hj², the lateral force at the top floor in kN is ____.
Numerical answer — type the value.
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Answer: 810
With equal weights, share ∝ h²: 9, 36, 81, which sum to 126. Top-floor force = 1260 × 81/126 = 810 kN. The other floors get 90 and 360 kN, and 90 + 360 + 810 = 1260.
A cable spans 40 m horizontally with a sag of 4 m and carries a load of 5 kN per metre of horizontal span. Using H = wL²/8d, the maximum tension at the supports in kN, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 269.3
H = 5 × 40²/(8 × 4) = 5 × 1600/32 = 250 kN. Vertical reaction V = wL/2 = 5 × 40/2 = 100 kN. Tmax = √(250² + 100²) = √72 500 = 269.26, which is 269.3 kN. Check: 269.3² ≈ 72 522 against 72 500 within rounding.